Question

Formatted question description: https://leetcode.ca/all/199.html

Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

Example 1:

Input: root = [1,2,3,null,5,null,4]
Output: [1,3,4]


Example 2:

Input: root = [1,null,3]
Output: [1,3]


Example 3:

Input: root = []
Output: []


Constraints:

• The number of nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

Algorithm

Just save the rightmost node of each level.

Code

• import java.util.ArrayList;
import java.util.List;
import java.util.Queue;

public class Binary_Tree_Right_Side_View {

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Integer> rightSideView(TreeNode root) {

List<Integer> result = new ArrayList<>();

if (root == null) {
return result;
}

q.offer(root);

int currentLevelCount = 1;
int nextLevelCount = 0;

while (!q.isEmpty()) {
TreeNode current = q.poll();
currentLevelCount--;

if (current.left != null) {
q.offer(current.left);
nextLevelCount++;
}
if (current.right != null) {
q.offer(current.right);
nextLevelCount++;
}

if (currentLevelCount == 0) {
currentLevelCount = nextLevelCount;
nextLevelCount = 0;
}
}

return result;
}
}
}

############

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> ans = new ArrayList<>();
if (root == null) {
return ans;
}
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
while (!q.isEmpty()) {
for (int n = q.size(); n > 0; --n) {
TreeNode node = q.poll();
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/binary-tree-right-side-view/
// Time: O(N)
// Space: O(H)
class Solution {
vector<int> ans;
void dfs(TreeNode *root, int lv) {
if (!root) return;
if (lv == ans.size()) ans.push_back(root->val);
else ans[lv] = root->val;
dfs(root->left, lv + 1);
dfs(root->right, lv + 1);
}
public:
vector<int> rightSideView(TreeNode* root) {
dfs(root, 0);
return ans;
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
ans = []
if root is None:
return ans
q = deque([root])
while q:
ans.append(q[-1].val)
for _ in range(len(q)):
node = q.popleft()
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
return ans

#############

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
from collections import deque

class Solution(object):
def rightSideView(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""

def dfs(root, h):
if root:
if h == len(ans):
ans.append(root.val)
# pre-order, all the way to the right
dfs(root.right, h + 1)
dfs(root.left, h + 1)

ans = []
dfs(root, 0)
return ans


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func rightSideView(root *TreeNode) (ans []int) {
if root == nil {
return
}
q := []*TreeNode{root}
for len(q) > 0 {
ans = append(ans, q[len(q)-1].Val)
for n := len(q); n > 0; n-- {
node := q[0]
q = q[1:]
if node.Left != nil {
q = append(q, node.Left)
}
if node.Right != nil {
q = append(q, node.Right)
}
}
}
return
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function rightSideView(root: TreeNode | null): number[] {
const ans = [];
if (!root) {
return ans;
}
const q = [root];
while (q.length) {
const n = q.length;
ans.push(q[n - 1].val);
for (let i = 0; i < n; ++i) {
const { left, right } = q.shift();
left && q.push(left);
right && q.push(right);
}
}
return ans;
}


• // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
use std::collections::VecDeque;
impl Solution {
pub fn right_side_view(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
let mut res = vec![];
if root.is_none() {
return res;
}
let mut q = VecDeque::new();
q.push_back(root);
while !q.is_empty() {
let n = q.len();
res.push(q[n - 1].as_ref().unwrap().borrow().val);
for _ in 0..n {
if let Some(node) = q.pop_front().unwrap() {
let mut node = node.borrow_mut();
if node.left.is_some() {
q.push_back(node.left.take());
}
if node.right.is_some() {
q.push_back(node.right.take());
}
}
}
}
res
}
}